# If He gas has an average kinetic energy of 4570 J/mol under certain conditions, what is the root...

## Question:

If He gas has an average kinetic energy of 4570 J/mol under certain conditions, what is the root mean square speed of {eq}N_2 {/eq} gas molecules under the same conditions?

## Kinetic Energy

In physics, kinetic energy is a form of energy due to the motion of an object or particle. It is proportional to the product of mass and square of velocity of the object or particle. Here, we are investigating the kinetic energy of a He and N2 gas.

Given:

• kinetic energy of He gas {eq}KE_{He} = 4570 \ \text{J}/\text{mol}{/eq}
• molar mass of N2 {eq}m_{N_2} = 0.028 \ \text{kg}/\text{mol}{/eq}
• gas constant {eq}R = 8.314 \ \text{J}/\text{mol K}{/eq}

Required:

• root mean square velocity of N2 {eq}v_{N_2} = ?{/eq}

First, we need to solve for the temperature.

{eq}KE_{He} = \frac{3}{2} R T\\ \Rightarrow T = \frac{(2)KE_{He}}{(3)R}\\ \Rightarrow T = \frac{(2)(4570 \ \text{J}/\text{mol})}{(3)(8.314 \ \text{J}/\text{mol K})}\\ \Rightarrow T = 366.45 \ \text{K} {/eq}

Now, we can solve for the root mean square velocity

{eq}v_{N_2} = \sqrt{\frac{3RT}{m_{N_2}}}\\ v_{N_2} = \sqrt{\frac{3(8.314 \ \text{J}/\text{mol K})( 366.45 \ \text{K})}{0.028 \ \text{kg}/\text{mol}}}\\ \boxed{v_{N_2} = 571.34 \ \text{m}/\text{s}} {/eq}